Question

What is the molarity of \[11.2\] \[V\] of \[{H_2}{O_2}\]\[?\]
A.\[1\] M
B.\[2\] M
C.\[5.6\] M
D.\[11.2\] M

Answer 1

Hint: First we know that the molarity (M) of a solution is the number of moles of solute dissolved in one litre of solution. The solution includes both the solute and the solvent. Hence the molarity of a solution is the ratio of the moles of solute to the volume of the solution expressed in litres.

Complete answer:
Given \[11.2\] \[V\] of \[{H_2}{O_2}\] which is the solute. Let \[n\]be the number of moles of solute \[{H_2}{O_2}\] in a solution and \[v\]be the volume of a solution.
We know that one litre of \[{H_2}{O_2}\]​ aqueous solution provides \[11.2\] litre of \[{O_2}\]​ at STP (Standard temperature and pressure).
Moles of \[{O_2}\]\[ = \dfrac{{11.2}}{{22.4}} = 0.5\]
Then the number of moles in \[{H_2}{O_2}\] is \[n = 0.5 \times 2\].
molarity of\[{H_2}{O_2}\](M)​​ \[ = \dfrac{n}{v} = 1\]M
Hence the correct option is (A) \[1\]M.

Note:
We can also solve this problem by using the relationship between normality and molarity.
i.e., \[Molarity = \dfrac{{Normality}}{n}\]---(1)
We know that one normality (\[N\]) of \[{H_2}{O_2}\] is \[5.6\] times the volume strength (\[V\]).
Then \[1\;V = \dfrac{1}{{5.6}}N\]
\[ \Rightarrow 11.2\;V = \dfrac{{11.2}}{{5.6}}N = 2N\]
We know that \[n = 2\] for \[{H_2}{O_2}\]
Then the equation (1) becomes
\[Molarity = \dfrac{2}{2} = 1M\]
 Also note that the volume is in litres of solution and not litres of solvent. At Standard temperature and pressure, one mole of any gas will occupy a volume of \[22.4\] litre.